Basic proof of units in a ring with identity.

In summary: Another way to think about it is to consider the equation ab=1 and note that since b is a unit, there exists an element b^-1 such that bb^-1=1. Then ab=b(b^-1a)=1, so a is also a unit. Therefore, both a and b are units. In summary, if ab is a unit in a ring with identity and neither a nor b is a zero divisor, then a and b are also units in the ring.
  • #1
shamus390
8
0

Homework Statement



Let R be a ring with identity, and a,b are elements in R. If ab is a unit, and neither a nor b is a zero divisor, prove a and b are units.


Homework Equations



If ab is a unit then (ab)c=1=c(ab) for some c in R.


The Attempt at a Solution


Assume both a and b are not zero divisors, and denote the identity element of R as 1.

Since ab is a unit in R, there exists some c such that (ab)c = 1.
(ab)c=1 [itex]\Rightarrow[/itex] (ca)b = 1 [itex]\Rightarrow[/itex] b is a unit.

Similarly, (ab)c= 1 [itex]\Rightarrow[/itex] a(bc) [itex]\Rightarrow[/itex] a is a unit.

Does the above sufficiently prove the claim in the problem statement? I feel like I'm missing something.

Thanks!
 
Physics news on Phys.org
  • #2
shamus390 said:

Homework Statement



Let R be a ring with identity, and a,b are elements in R. If ab is a unit, and neither a nor b is a zero divisor, prove a and b are units.

Homework Equations



If ab is a unit then (ab)c=1=c(ab) for some c in R.

The Attempt at a Solution


Assume both a and b are not zero divisors, and denote the identity element of R as 1.

Since ab is a unit in R, there exists some c such that (ab)c = 1.
(ab)c=1 [itex]\Rightarrow[/itex] (ca)b = 1 [itex]\Rightarrow[/itex] b is a unit.

Similarly, (ab)c= 1 [itex]\Rightarrow[/itex] a(bc) [itex]\Rightarrow[/itex] a is a unit.

Does the above sufficiently prove the claim in the problem statement? I feel like I'm missing something.

Thanks!

Yes, you should feel uncomfortable with that. You didn't use that a and b weren't zero divisors. If (ab)c=1 then you can regroup that to a(bc)=1. That doesn't prove a is a unit. You also need to show (bc)a=1. Don't assume the ring is commutative.
 
  • #3
Dick said:
Yes, you should feel uncomfortable with that. You didn't use that a and b weren't zero divisors. If (ab)c=1 then you can regroup that to a(bc)=1. That doesn't prove a is a unit. You also need to show (bc)a=1. Don't assume the ring is commutative.

Ah thank you, that was a foolish oversight, they wouldn't stipulate a and b weren't zero divisors if it wasn't necessary...

So if we let a, b and c be elements of R such that ab = ac, then:

ab-ac=0R = a(b-c) = 0R, and
a is not a zero divisor [itex]\Rightarrow[/itex] (b - c) = 0R [itex]\Rightarrow[/itex] (b-c) = 0R [itex]\Rightarrow[/itex] b = c

Does that allow me to claim the following?

Multiplying a(bc) = 1 by a to give a(bc)a = 1a , (1a=a1 by definition) then, a(bc)a = a1 and by cancellation, (bc)a = 1 [itex]\Rightarrow[/itex] a is a unit.
 
  • #4
shamus390 said:
Ah thank you, that was a foolish oversight, they wouldn't stipulate a and b weren't zero divisors if it wasn't necessary...

So if we let a, b and c be elements of R such that ab = ac, then:

ab-ac=0R = a(b-c) = 0R, and
a is not a zero divisor [itex]\Rightarrow[/itex] (b - c) = 0R [itex]\Rightarrow[/itex] (b-c) = 0R [itex]\Rightarrow[/itex] b = c

Does that allow me to claim the following?

Multiplying a(bc) = 1 by a to give a(bc)a = 1a , (1a=a1 by definition) then, a(bc)a = a1 and by cancellation, (bc)a = 1 [itex]\Rightarrow[/itex] a is a unit.

Yes, that's the idea.
 

1. What is a unit in a ring with identity?

A unit in a ring with identity is an element that has a multiplicative inverse. In other words, for an element to be a unit, there must exist another element in the ring that, when multiplied with the original element, equals the identity element (usually denoted as 1).

2. Why is it important to prove the existence of units in a ring with identity?

Proving the existence of units in a ring with identity is important because it helps establish the fundamental properties of the ring, such as the distributive law and the existence of a multiplicative identity. It also allows for the use of algebraic techniques to solve equations and prove theorems.

3. How is the existence of units proven in a ring with identity?

The existence of units in a ring with identity is proven by showing that every non-zero element in the ring has a multiplicative inverse. This can be done by explicitly finding the inverse for each element or by showing that the product of an element and its inverse equals the identity element.

4. Can there be more than one unit in a ring with identity?

No, there can only be one unit in a ring with identity. This is because the unit element must be unique and multiplying any other element with it will not produce a different result.

5. What is the significance of units in a ring with identity in real-world applications?

Units in a ring with identity have many practical applications in mathematics, physics, and computer science. They are used to solve equations and prove theorems in abstract algebra, and they also have applications in coding theory and cryptography. In physics, units in a ring with identity are used to model physical phenomena and make predictions about the behavior of systems.

Similar threads

Proving Closure and Identity for U(n)
    • Calculus and Beyond Homework Help
Replies
16
Views
4K
Abstract Algebra: Ring Proof (Multiplicative Inverse)
    • Calculus and Beyond Homework Help
Replies
1
Views
1K
Prove ##(a\cdot b)\cdot c =a\cdot (b \cdot c)## using Peano postulates
    • Calculus and Beyond Homework Help
Replies
1
Views
460
Prove ##1 + a=s(a)=a+1## for ##a \in \mathbb{N}##
    • Calculus and Beyond Homework Help
Replies
1
Views
578
Prove if ##a\cdot c = b \cdot c## then ##a = b## using Peano postulates
    • Calculus and Beyond Homework Help
Replies
2
Views
274
Prove ##(a+b)\cdot c=a\cdot c+b\cdot c## using Peano postulates
    • Calculus and Beyond Homework Help
Replies
3
Views
521
Showing that an element is a unit in a ring
    • Calculus and Beyond Homework Help
Replies
3
Views
1K
Polynomial ring zero divisors
    • Calculus and Beyond Homework Help
Replies
13
Views
2K
Extending an Additive Group Homomorph. to a Ring Homomorph.
    • Calculus and Beyond Homework Help
Replies
1
Views
1K
An alternative proof (Hopefully not an alternative fact)
    • Calculus and Beyond Homework Help
Replies
2
Views
904

Hot Threads

Recent Insights

Back
Top